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I'd like to derive an asymptotic expression for the following function (as $m \rightarrow \infty$) involving the incomplete Beta function:

$$ F(m) = \frac{1}{2} B_{1/2}(m, m) - B_{1/2}(m+1, m) $$

where $m$ is a non-negative integer.

I've attempted to use Laplace's method, but the contributions I get from each of the terms cancel out.

Could anyone suggest an alternative approach? Is there perhaps a way of getting higher order asymptotic terms from Laplace's method?


The definition I am working from for the incomplete Beta function is:

$$ B_z(a,b) := \int_0^z \alpha^{a-1} (1-\alpha)^{b-1} d\alpha.$$

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1 Answer 1

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We have

$$F(m)=\frac{1}{2} B_{1/2}(m, m) - B_{1/2}(m+1, m)\qquad\qquad\qquad\qquad\qquad\qquad\ \ $$

$$=\frac{1}{2}\left[\int_0^{1/2} x^{m-1} (1-x)^{m-1} dx-2\int_0^{1/2} x^{m} (1-x)^{m-1} dx\right]$$

$$=\frac{1}{2}\int_0^{1/2}x^{m-1}(1-x)^{m-1}(1-2x)dx\qquad\qquad\qquad\qquad\quad$$

$$=\frac{1}{2}\int_0^{1/2}[x^{m-1}(1-x)^{m}-x^{m}(1-x)^{m-1}]dx\qquad\qquad\qquad$$

$$=\frac{1}{2}\left[\frac{z^m(1-z)^m}{m}\right]_0^{1/2}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\ \ $$

$$=\frac{1}{m{2}^{1+2m}}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ $$

which has limit zero as $m\to\infty$.


The working shows that for any $z\in[0,1]$:

$$\frac{1}{2}B_z(m,m)-B_z(m+1,m)=\frac{z^m(1-z)^m}{2m},$$

and for any $z\in[0,1]$ the limit as $m\to\infty$ is zero.

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  • $\begingroup$ Nicely done! +1 $\endgroup$
    – Mark Viola
    Aug 5, 2016 at 14:44
  • $\begingroup$ Nice indeed, good job $\endgroup$
    – Yuriy S
    Aug 5, 2016 at 15:29
  • $\begingroup$ Wow, that's a really neat solution. I really wasn't expecting an exact answer! Thanks so much! :) $\endgroup$
    – Tariq
    Aug 5, 2016 at 17:21

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